Question: Rationalize the denominator of $\displaystyle \frac{1}{\sqrt[3]{3} - \sqrt[3]{2}}$. With your answer in the form $\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}$, and the fraction in lowest terms, what is $A + B + C + D$?
Since the denominator involves cube roots, we can't just multiply by a conjugate. Instead we use the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$. Letting $a = \sqrt[3]{3}$ and $b = \sqrt[3]{2}$, we have \[
\frac{1}{\sqrt[3]{3} - \sqrt[3]{2}} = \frac{1}{\sqrt[3]{3} - \sqrt[3]{2}} \cdot \frac{(\sqrt[3]{3})^2 + \sqrt[3]{3} \sqrt[3]{2} + (\sqrt[3]{2})^2}{(\sqrt[3]{3})^2 + \sqrt[3]{3} \sqrt[3]{2} + (\sqrt[3]{2})^2}.
\]The denominator simplifies by the above identity to $(\sqrt[3]{3})^3 - (\sqrt[3]{2})^3 = 1$, so we are left with $\sqrt[3]{9} + \sqrt[3]{6} + \sqrt[3]{4}$. Matching the form given in the problem, $D = 1$ and $A = 9$, $B = 6$, $C = 4$ (in some order), so $A+B+C+D = \boxed{20}$.